If the sum of n consecutive odd numbers is ${25}^2-{11}^2$, then 

Let n consecutive odd numbers be $2\mathrm{k}+1,2\mathrm{k}+3,2\mathrm{k}+5,\dots ,2\mathrm{k}+2\mathrm{n}-1$
According to question, sum of these n numbers  
$\begin{array}{l}=\frac{\mathrm{n}}{2}[2\mathrm{k}+1+2\mathrm{k}+2\mathrm{n}-1]=\mathrm{n}(2\mathrm{k}+\mathrm{n})\\ ={\mathrm{n}}^2+2\mathrm{kn}=(\mathrm{n}+\mathrm{k}{)}^2-{\mathrm{k}}^2\end{array}$
Given that, $(\mathrm{n}+\mathrm{k}{)}^2-{\mathrm{k}}^2={25}^2-{11}^2$
$\Rightarrow \mathrm{n}+\mathrm{k}=25\text{ and }\mathrm{k}=11\Rightarrow \mathrm{n}=14\text{ and }\mathrm{k}=11$
So, first term $=2\mathrm{k}+1=23$
last term $=2\mathrm{k}+2\mathrm{n}-1=22+28-1=22+27=49$